1. ## a better factorization?

Hey,
I need to factor this to apply partial fractions method to a problem I'm working on. I came up with a solution, but I've always been bad at factorization and I was wondering if someone could come up with a better one.
Here is the polynomial:
$
-8x^4 + 4x^3 + 6x^2 - 5x + 1
$

here is the decomposition I came up with:
$-8x^4 + 4x^3 + 6x^2 - 5x + 1$
$= (-8x^4 + 4x^3) + (6x^2 - 5x + 1)$
$= x^3(x - 1/2) + (x - 1/3)(x - 1/2)$
$= (x - 1/2)(x^3 + x - 1/3)$

Thanks!

2. Originally Posted by epsilon
Hey,
I need to factor this to apply partial fractions method to a problem I'm working on. I came up with a solution, but I've always been bad at factorization and I was wondering if someone could come up with a better one.
Here is the polynomial:
$
-8x^4 + 4x^3 + 6x^2 - 5x + 1
$

here is the decomposition I came up with:
$-8x^4 + 4x^3 + 6x^2 - 5x + 1$
$= (-8x^4 + 4x^3) + (6x^2 - 5x + 1)$
$= x^3(x - 1/2) + (x - 1/3)(x - 1/2)$
$= (x - 1/2)(x^3 + x - 1/3)$

Thanks!
i've got a better way to solve this problem..
$-8x^4 + 4x^3 + 6x^2 - 5x + 1$
$-(8x^4-4x^3)+(3x-1)(2x-1)$
$-4x^3(2x-1)+(3x-1)(2x-1)$
$-(2x-1)[4x^3-(3x-1)]$
$-(2x-1)(4x^3-3x+1)$

ok, now we have to use try and error method to find another one root..

$(4x^3-3x+1)=0$
let's plug in $x=-1$
$4(-1)^3-3(-1)+1=0$
LHS=RHS
so -1 is the another root or $(x-1)$
then perform long division
$(x-1)/(4x^3-3x+1)$
...
...
$-(2x-1)(x-1)(4x^2-4x+1)$
$-(2x-1)(x-1)(2x-1)^2$